Question

Show that a function \( f: R_{+} \to A \subset N \), defined as \( f(x) = 4x^2 + 12x + 15 \) is one-one. Find set \( A \) so that \( f \) is onto where \( R_{+} = [0, \infty) \). Also, find if there exists \( a \in R_{+} \) such that \( f(a) = 7 \). Justify.

Hint:

Write \( f(x) = (2x + 3)^2 + 6 \) and use \( x \geq 0 \) to get \( f(x) \geq 15 \). The range \( [15, \infty) \) gives A, and 7 is below it.

Answer 1

Step 1: Set up the one-one test:
A function is one-one if \( f(x_1) = f(x_2) \) forces \( x_1 = x_2 \). Here the domain is \( R_{+} = [0, \infty) \), so take \( x_1, x_2 \geq 0 \) with \( f(x_1) = f(x_2) \).
\[ 4x_1^2 + 12x_1 + 15 = 4x_2^2 + 12x_2 + 15 \]

Step 2: Factorise and conclude:
Subtract and collect the terms.
\[ 4(x_1^2 - x_2^2) + 12(x_1 - x_2) = 0 \]
\[ (x_1 - x_2)\,[\,4(x_1 + x_2) + 12\,] = 0 \]
Both \( x_1 \) and \( x_2 \) are non-negative, so \( 4(x_1 + x_2) + 12 \geq 12 \). This factor is never zero.
So \( x_1 - x_2 = 0 \), that is \( x_1 = x_2 \). Hence \( f \) is one-one.

Step 3: Find the range of f:
For \( f \) to be onto, the set \( A \) must be exactly the range of \( f \). Complete the square.
\[ f(x) = 4x^2 + 12x + 15 = (2x + 3)^2 + 6 \]
For \( x \geq 0 \) we have \( 2x + 3 \geq 3 \), so \( (2x + 3)^2 \geq 9 \).
So \( f(x) \geq 15 \), and \( f(0) = 15 \).

Step 4: Show every value from 15 upward is reached:
Put \( y = (2x + 3)^2 + 6 \) with \( y \geq 15 \). Then \( 2x + 3 = \sqrt{y - 6} \), so \( x = \dfrac{\sqrt{y - 6} - 3}{2} \).
For \( y \geq 15 \) we get \( \sqrt{y - 6} \geq 3 \), so \( x \geq 0 \). Every \( y \geq 15 \) has a pre-image in \( R_{+} \).
So the range is \( [15, \infty) \) and the set for which \( f \) is onto is \( A = [15, \infty) \).

Step 5: Check whether f(a) = 7 is possible:
Every value of \( f \) on \( R_{+} \) is at least 15, and \( 7 \lt 15 \). So \( 7 \) is not in the range.
Check by solving the equation. \( 4a^2 + 12a + 15 = 7 \) gives \( 4a^2 + 12a + 8 = 0 \).
\[ a^2 + 3a + 2 = 0 \Rightarrow (a + 1)(a + 2) = 0 \Rightarrow a = -1 \text{ or } a = -2 \]
Both values are negative, so neither lies in \( R_{+} \).

Final Answer:
\( f \) is one-one, \( A = [15, \infty) \), and there is no \( a \in R_{+} \) with \( f(a) = 7 \), because \( f(x) \geq 15 \) for all \( x \geq 0 \). \[ \boxed{A = [15, \infty),\ \text{no such } a \text{ exists}} \]